2023～2024学年度第二学期期末学业水平诊断高二数学参考答案及评分标准一、选择题CCADBDCA二、选择题9.ABD10.BCD11.AC三、填空题1423−80(−∞,]()n−1LL212.13.e14.3，45四、解答题15.解：（1）根据已知条件，可得：获奖没有获奖合计选修阅读课程81220不选阅读课程22830合计104050······················································3分零假设为H：创新作文比赛获奖与选修阅读课程无关联，0根据列联表中数据计算得到，50×(8×28−2×12)225χ2==≈8.333>7.879.·······························6分20×30×10×403根据小概率值α=0.005的独立性检验，推断H不成立，即认为创新作文比赛获奖与0选修阅读课程有关联，此推断犯错误的概率不大于0.005.····························7分（2）由题意可知X的可能取值为1,2,3,则···································8分C1C21C2C17P(X=1)=82=，P(X=2)=8=2，C315C3151010C37P(X=3)=8=，········································11分C31510所以,随机变量X的分布列为:X123177P15151517712E(X)=×1+2×+3×=.··························13分所以151515516.解：（1）当a=−2时，f(x)=(x2−2x+1)ex，所以f′(=x)(x2−1)ex.·········1分(x,y)y=(x2−2x+1)ex=k(x2−1)ex设切点为，则0，0，000000高二数学答案（，）y−(x2−2x+1)ex=(x2−1)ex(x−x)所以，切线方程为00.························3分0000将(1,0)代入得(x−1)2x=0，解得x=0或x=1.·····························5分0000故过(1,0)的切线方程为y=0或x+y−1=0.················································7分（2）f′(x)=(2x+a)ex+(x2+ax+1)ex=(x+a+1)(x+1)ex.·····················8分当a=0时，f′(x=)(x+1)2ex，恒有f′(x)≥0，函数f(x)单调递增.·········10分当a>0时，−a−1<−1，当x∈(−∞,−a−1)，或x∈(−1,+∞)时，f′(x)>0，函数f(x)单调递增，当x∈(−a−1,−1)时，f′(x)<0，函数f(x)单调递减.····12分当a<0时，−a−1>−1，当x∈(−∞,−1)，或x∈(−a−1,+∞)时，f′(x)>0，函数f(x)单调递增，当x∈(−1,−a−1)时，f′(x)<0，函数f(x)单调递减.·······14分综上，当a=0时，f(x)在R上单调递增，当a>0时，f(x)在(−∞,−a−1)，(−1,+∞)上单调递增，在(−a−1,−1)上单调递减，当a<0时，f(x)在(−∞,−1)，(−a−1,+∞)上单调递增，在(−1,−a−1)上单调递减.······························15分17.解：（1）由题意可知，b−b=a，即b−1=−1，故b=0.························1分21222由b−b=a，可得a=1.······················································2分3233所以数列{a}的公差d=2，所以a=−1+2(n−2)=2n−5.······················3分nn由b−b=a，b−b=a，，b−b=a，nn−1nn−1n−2n−1212(n−1)(−1+2n−5)叠加可得b−b=a+a++a=，n123n2整理可得b=n2−4n+4(n≥2)；当n=1时，满足上式，n所以b=n2−4n+4················································································5分n(n−2)2+5（2）不妨设